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16x^2+56x+5=0
a = 16; b = 56; c = +5;
Δ = b2-4ac
Δ = 562-4·16·5
Δ = 2816
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2816}=\sqrt{256*11}=\sqrt{256}*\sqrt{11}=16\sqrt{11}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(56)-16\sqrt{11}}{2*16}=\frac{-56-16\sqrt{11}}{32} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(56)+16\sqrt{11}}{2*16}=\frac{-56+16\sqrt{11}}{32} $
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